Joel Kolstad wrote:
On a reciever, I'm using a MMIC LNA at VHF (2m), and while it's quite clear to
me why I need to match at the amplifier's input -- to minimize VSWR and get as
much of the signal from the antenna into the system -- it's not quite to clear
to me just how important output matching is. The S22 parameters for this LNA
are very close to an open circuit with a little capacitance -- I expect the
output looks back into a drain or something else that's a reasonably decent
current source at RF, and you see that shunted by parasitic capacitances.

So... how important is it that I bother to build a matching network at the
output? (I'll be "outputting" to another ~50 ohm input.) It doesn't really
seem that important, and building something like an L-match with de-Q-ing
resistor drops the gain a couple dB anyway.

I know that for a transmitter output matching is important so that reflections
from poor or non-existant terminations don't blow up the output amplifier, but
for low-level signals as in a receiver it wouldn't seem to matter so much?

Thanks,
---Joel Kolstad

OK, since you called it an LNA, I assume you want to take advantage of
the low noise. In that case, you do NOT necessarily want to match the
input for low SWR. You want to match to the amplifier's input noise
resistance: the equivalent input noise voltage divided by the
equivalent input noise current. This in general is NOT the same as the
input impedance. Adjust the input match for the lowest noise figure,
not the lowest SWR, if it's low noise you want.

Output matching will transfer the greatest power to the load. Assuming
S12 is very low, adjusting the output matching will not materially
affect the input power, so maximizing the output power will maximize
the gain. You may or may not have a need to do that. And you need to
pay attention to amplifier stability: is it unconditionally stable, or
must you keep the load within some bounds to keep it stable? Have a
look for articles about "maximum available gain" and "maximum usable
gain." For example, see
http://www.microwaves101.com/encyclo...factor.cfm#mag.

Cheers,
Tom

Hi Tom,

Thanks for the information; someone else pointed me yesterday to the idea that
you'd generally match the input for optimum noise performance rather than
optimum power transfer. The data sheet claims that the optimum noise figure
is 0.8dB, but the part itself is marketed as "typical 1.8dB NF" -- which they
obtained from their own eval board that, as far as I can tell, was matched
pretty close to optimum power transfer. In any case, at present mine is also
matched at the input for optimum power transfer, and I'll measure its noise
figure and see whether or not I'm around 1.8dB or better... if so I don't
think I'm going to worry about it too much.

Output matching will transfer the greatest power to the load.

Yeah, but it's not practical, is it? I'm looking back at (approximately) a
current source, and my load is dictated as (approximately) 50 ohms, so the
load itself is what's driving the power transferred, no?

Assuming
S12 is very low

It is, 0.01 everywhere.

And you need to
pay attention to amplifier stability: is it unconditionally stable, or
must you keep the load within some bounds to keep it stable?

I haven't calculated its k factor; I'll do that soon -- thanks for the link.

---Joel

Joel Kolstad wrote:
Hi Tom,

Thanks for the information; someone else pointed me yesterday to the idea that
you'd generally match the input for optimum noise performance rather than
optimum power transfer. The data sheet claims that the optimum noise figure
is 0.8dB, but the part itself is marketed as "typical 1.8dB NF" -- which they
obtained from their own eval board that, as far as I can tell, was matched
pretty close to optimum power transfer. In any case, at present mine is also
matched at the input for optimum power transfer, and I'll measure its noise
figure and see whether or not I'm around 1.8dB or better... if so I don't
think I'm going to worry about it too much.

Output matching will transfer the greatest power to the load.

Yeah, but it's not practical, is it? I'm looking back at (approximately) a
current source, and my load is dictated as (approximately) 50 ohms, so the
load itself is what's driving the power transferred, no?

Well, no... Clearly it's not exactly a current source. Perhaps it's
1000 ohms, or 10000 ohms, (plus some reactance, of course) which is
high but not a pure current source. Then a network that matches 1000
ohms, or 10000 ohms, to 50 ohms (and cancels the reactance), will give
you the most power output. That is a linear, small-signal model, but
that should be a pretty good approximation for any application where
you need a low-noise preamp.

You do need to consider losses in whatever matching network you use;
and many matching networks will be highly resonant to transform between
impedances that are in a large ratio. Realize that the match at the
operating frequency may give you a load at some other frequency which
causes instability...

If the source (the output impedance) was really a current source, you
could get near-infinite power gain (again, assuming that S12 is zero)
by transforming the load to as high an impedance as possible. Consider
the small-signal low-frequency model of a bipolar transistor where the
output is a current source shunted by a resistance. Assume that
(internal effective) resistance is infinite. Now the current from the
current source all flows into the load. Since power is i^2*R,
increasing R increases the power without bound. A matching network
just transforms your practical load (e.g. 50 ohms) to the desired load
R.

What's practical depends on how wide a bandwidth you want, and how good
you are at designing and building impedance matching networks. (I can
personally come up with lots of IMpractical networks! ;-)

Cheers,
Tom

Thanks again Tom, you've really cleared things up for me here.

Here's the concrete numbers... in the center of the band, S22 -- converted
back to an impedance -- is 119.5-j90 ohms. With a 2V source, if I conjugate
match my 50 ohm load obviously I get 1V across the resistor or 20mW = 13dBm.
Without a match... let's see... current around the loop is 2/(119.5-j90+50) =
9.2+j4.89 mA, and power in the 50 ohm load is I^2*R = 3.04+j4.5mW = (looking
just at real power) 4.83dBm. So... without a match... I'd be leaving 13-4.83
= 8.18dB of gain "on the floor"... ouch!

---Joel

To chime in late: typically the routine in LNA design is noise match
at the input, then conjugate match at the output considering the
effects of the noise match. Really the only reason to play games with
the input noise match is either for broadbanding or to address a
specific stability issue, which usually is better handled by judicious
feedback.


Joel Kolstad wrote:
Thanks again Tom, you've really cleared things up for me here.

Here's the concrete numbers... in the center of the band, S22 -- converted
back to an impedance -- is 119.5-j90 ohms. With a 2V source, if I conjugate
match my 50 ohm load obviously I get 1V across the resistor or 20mW = 13dBm.
Without a match... let's see... current around the loop is 2/(119.5-j90+50) =
9.2+j4.89 mA, and power in the 50 ohm load is I^2*R = 3.04+j4.5mW = (looking
just at real power) 4.83dBm. So... without a match... I'd be leaving 13-4.83
= 8.18dB of gain "on the floor"... ouch!

---Joel

On Oct 25, 2:50 pm, "Joel Kolstad"
wrote:
Thanks again Tom, you've really cleared things up for me here.

Here's the concrete numbers... in the center of the band, S22 -- converted
back to an impedance -- is 119.5-j90 ohms. With a 2V source, if I conjugate
match my 50 ohm load obviously I get 1V across the resistor or 20mW = 13dBm.
Without a match... let's see... current around the loop is 2/(119.5-j90+50) =
9.2+j4.89 mA, and power in the 50 ohm load is I^2*R = 3.04+j4.5mW = (looking
just at real power) 4.83dBm. So... without a match... I'd be leaving 13-4.83
= 8.18dB of gain "on the floor"... ouch!

---Joel

OK, things aren't quite as bad as you've painted them, Joel. Given the
analysis you did for the second half, what you need to do for the first
half is to look at the power delivered to 119.5 ohms with 1 volt across
it. That's 1/119.5 watts, or about 8.4 milliwatts. That's about
+9.23dBm That's the maximum power you can get from that source. Into
50 ohms with no matching, the magnitude of the current is 10.42mA.
The resistor does NOT care what phase angle relative to your arbitrary
reference that is. So the power in that case is 5.43mW, or +7.35dBm.
You've left not 8dB, but just under 2dB, of gain "on the floor."
Things don't get really bad till the mismatch is quite a bit worse than
that. Note that your 2dB occurs with an SWR of about 4:1. At 2:1 SWR,
you only lose half a dB.

Cheers,
Tom

Thanks for the correction, Tom. They really did give me a EE at one point in
time, believe it or not, it's just been forever since I've had to do Real
Work!
---Joel

K7ITM a écrit :
...

What's practical depends on how wide a bandwidth you want, and how good
you are at designing and building impedance matching networks. (I can
personally come up with lots of IMpractical networks! ;-)

1Meg is a pretty high value so we all can conclude that you're excellent :-)


--
Thanks,
Fred.